Finite Dimensional Subspace of Normed Vector Space is Closed

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Theorem

Let $V$ be a normed vector space.

Let $W$ be a finite dimensional subspace of $V$.

And the ground field $K$ is complete.


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Then $W$ is closed.



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Proof 1

Suppose that $\dim W = n$.

Let:

$\set {e_1, e_2, \ldots, e_n}$

be a basis for $W$.

Aiming for a contradiction, suppose that $W$ is not a closed set.

Then there exists a convergent sequence $\sequence {w_k}$ in $W$ such that:

$w_k \to w$ in $V$

where $w \in V \setminus W$.

Note that:

$\set {e_1, e_2, \ldots, e_n}$

is linearly independent in $W$, and hence $V$.

Note that since $w \not \in W$, $w$ cannot be written as a linear combination of elements of $\set {e_1, e_2, \ldots, e_n}$.

So:

$\set {e_1, e_2, \ldots, e_n, w}$

is linearly independent in $V$.


So consider the subspace:

$W^* = \span \set {e_1, e_2, \ldots, e_n, w}$.

Using the sequence $\sequence {w_n}$ from before, write:

$w_k = \tuple {w_k^{\paren 1}, w_k^{\paren 2}, \ldots, w_k^{\paren n}, 0} \in W^*$

and:

$w = \tuple {0, 0, \ldots, 0, 1} \in W^*$

Since Norms on Finite-Dimensional Real Vector Space are Equivalent, $w_k\to w$ in direct product norm.

Since sequence of vectors converges iff each component converges, we necessarily have:

$w_k^{\paren j} \to 0$

for each $1 \le j \le n$.

However, since the $(n+1)$-th component of $w_k$ is 0 and that of $w$ is $1$, we would also have:

$0 \to 1$

Clearly this is impossible, so we have derived a contradiction.

So $W$ is necessarily closed.

$\blacksquare$


Proof 2

Because $K$ is complete, the finite dimensional normed space $W$ is complete.

By Subspace of Complete Metric Space is Closed iff Complete, $W$ is closed.

$\blacksquare$


Necessity of completeness of the ground field


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$K = \Q$ is not complete.

$W = \Q$

$V = \Q \sqbrk {\sqrt 2}$

Since

$\dim_K W = 1$

$W$ is a finite dimensional subspace of $V$.

But $W$ is not closed in $V$.


Also see


Source

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